As far as I know, $K(\pi,1)$ space is a manifold $M$ such that $\pi_n(M) = 0$ for $n > 1$ and $\pi_1(M) = \pi$.
Q. Why is the singular cohomology $H_{\mathrm{sing}}^i(M, {\Bbb Z}/n)$ isomorphic to the group cohomology group $H_{\mathrm{group}}^i(\pi, {\Bbb Z}/n)$?
Most likely one can generalise this notion to schemes, in which case the singular cohomology must be replaced with the etale cohomology and the fundamental group with the Grothendieck's algebraic fundamental group.
If $M$ is a $K(\pi,1)$ space, its universal cover $\tilde M$ is contractible. Consider a simplicial decomposition of $M$ used to compute singular (co)homology it can be lifted to a simplicial decomposition of $\tilde M$ invariant by $\pi$, and this implies that it defines a free resolution of $\pi$-modules since $\tilde M$ is contractible.